This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 1049

2017 Puerto Rico Team Selection Test, 3

In triangle $ABC$, the altitude through $B$ intersects $AC$ at $E$ and the altitude through $C$ intersects $AB$ at $F$. Point $T$ is such that $AETF$ is a parallelogram and points $ A$ ,$T$ lie on different half-planes wrt the line $EF$. Point $D$ is such that $ABDC$ is a parallelogram and points $ A$ ,$D$ lie in different half-planes wrt line $BC$. Prove that $T, D$ and the orthocenter of $ABC$ are collinear.

1998 AIME Problems, 3

The graph of $y^2+2xy+40|x|=400$ partitions the plane into several regions. What is the area of the bounded region?

2020 Latvia TST, 1.4

It is given isosceles triangle $ABC$ with $AB = AC$. $AD$ is diameter of circumcircle of triangle $ABC$. On the side $BC$ is chosen point $E$. On the sides $AC, AB$ there are points $F, G$ respectively such that $AFEG$ is parallelogram. Prove that $DE$ is perpendicular to $FG$.

2010 Germany Team Selection Test, 2

Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P\subset R$ we have \[\frac{|R|}{|P|}\leq \sqrt 2\] where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively. [i]Proposed by Witold Szczechla, Poland[/i]

2012 France Team Selection Test, 2

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.

2013 APMO, 1

Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the center of its circumcircle. Show that the segments $OA$, $OF$, $OB$, $OD$, $OC$, $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.

1993 Moldova Team Selection Test, 8

Inside the parallelogram $ABCD$ points $M, N, K$ and $L{}$ are on sides $AB, BC, CD{}$ and $DA$, respectively. Let $O_1, O_2, O_3$ and $O_4$ be the circumcenters of triangles repesctively $MBN, NCK, KDL$ and $LAM{}$. Prove that the quadrilateral $O_1O_2O_3O_4$ is a parallelogram.

2014 Junior Balkan MO, 2

Consider an acute triangle $ABC$ of area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$, respectively $MND$. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.

2012 CentroAmerican, 3

Let $ABC$ be a triangle with $AB < BC$, and let $E$ and $F$ be points in $AC$ and $AB$ such that $BF = BC = CE$, both on the same halfplane as $A$ with respect to $BC$. Let $G$ be the intersection of $BE$ and $CF$. Let $H$ be a point in the parallel through $G$ to $AC$ such that $HG = AF$ (with $H$ and $C$ in opposite halfplanes with respect to $BG$). Show that $\angle EHG = \frac{\angle BAC}{2}$.

2007 Austria Beginners' Competition, 4

Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.

2010 Contests, 3

A rectangle formed by the lines of checkered paper is divided into figures of three kinds: isosceles right triangles (1) with base of two units, squares (2) with unit side, and parallelograms (3) formed by two sides and two diagonals of unit squares (figures may be oriented in any way). Prove that the number of figures of the third kind is even. [img]http://up.iranblog.com/Files7/dda310bab8b6455f90ce.jpg[/img]

2010 IMO Shortlist, 5

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

2005 International Zhautykov Olympiad, 2

Let the circle $ (I; r)$ be inscribed in the triangle $ ABC$. Let $ D$ be the point of contact of this circle with $ BC$. Let $ E$ and $ F$ be the midpoints of $ BC$ and $ AD$, respectively. Prove that the three points $ I$, $ E$, $ F$ are collinear.

2006 Moldova National Olympiad, 10.7

Consider an octogon with equal angles and rational side lengths. Prove that it has a symmetry center.

1991 Iran MO (2nd round), 2

Let $ABCD$ be a tetragonal. [b](a)[/b] If the plane $(P)$ cuts $ABCD,$ find the necessary and sufficient condition such that the area formed from the intersection of the plane $(P)$ and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case. [b](b)[/b] Consider one of the solutions of [b](a)[/b]. Find the situation of the plane $(P)$ for which the parallelogram has maximum area. [b](c)[/b] Find a plane $(P)$ for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of $ABCD.$

2017 BMT Spring, 10

Let $S$ be the set of points $A$ in the Cartesian plane such that the four points $A$, $(2, 3)$, $(-1, 0)$, and $(0, 6)$ form the vertices of a parallelogram. Let $P$ be the convex polygon whose vertices are the points in $S$. What is the area of $P$?

2005 All-Russian Olympiad, 1

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

2011 ISI B.Stat Entrance Exam, 5

$ABCD$ is a trapezium such that $AB\parallel DC$ and $\frac{AB}{DC}=\alpha >1$. Suppose $P$ and $Q$ are points on $AC$ and $BD$ respectively, such that \[\frac{AP}{AC}=\frac{BQ}{BD}=\frac{\alpha -1}{\alpha+1}\] Prove that $PQCD$ is a parallelogram.

2008 Sharygin Geometry Olympiad, 17

(A.Myakishev, 9--11) Given triangle $ ABC$ and a ruler with two marked intervals equal to $ AC$ and $ BC$. By this ruler only, find the incenter of the triangle formed by medial lines of triangle $ ABC$.

2018 Polish Junior MO First Round, 2

Inside parallelogram $ABCD$ is point $P$, such that $PC = BC$. Show that line $BP$ is perpendicular to line which connects middles of sides of line segments $AP$ and $CD$.

2010 Iran Team Selection Test, 6

Let $M$ be an arbitrary point on side $BC$ of triangle $ABC$. $W$ is a circle which is tangent to $AB$ and $BM$ at $T$ and $K$ and is tangent to circumcircle of $AMC$ at $P$. Prove that if $TK||AM$, circumcircles of $APT$ and $KPC$ are tangent together.

1991 India Regional Mathematical Olympiad, 5

Take any point $P_1$ on the side $BC$ of a triangle $ABC$ and draw the following chain of lines: $P_1P_2$ parallel to $AC$; $P_2P_3$ parallel to $BC$; $P_3P_4$ parallel to $AB$ ; $P_4P_5$ parallel to $CA$; and $P_5P_6$ parallel to $BC$, Here, $P_2,P_5$ lie on $AB$; $P_3,P_6$ lie on $CA$ and $P_4$ on $BC$> Show that $P_6P_1$ is parallel to $AB$.

1997 Tournament Of Towns, (528) 5

$E$ is the midpoint of the side $AD$ of a parallelogram $ABCD$. $F$ is the foot of the perpendicular from the vertex $B$ to the line $CE$. Prove that $ABF$ is an isosceles triangle. (MA Bolchkevich)

2006 China Team Selection Test, 1

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$, and let $L$ and $N$ be points on the side $AC$. The point $K$ is between $M$ and $B$, and the point $L$ is between $N$ and $C$. If $\frac{BK}{KM}=\frac{CL}{LN}$, then prove that the orthocentres of the triangles $\triangle{ABC}$, $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

2015 ELMO Problems, 3

Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. Let $X$ be the reflection of $A$ across the point $B$, and denote by $\gamma$ the circumcircle of triangle $BXC$. Suppose $\gamma$ and $\omega$ meet at $D \neq B$ and line $CD$ intersects $\omega$ at $E \neq D$. Prove that line $EX$ is tangent to the circle $\gamma$. [i]Proposed by David Stoner[/i]